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Following up on what has already been said, this is an explanation taken from "Group cohomology and gauge equivalence of some twisted quantum doubles", by Geoffrey Mason and Siu-Hung Ng.

Let $G$ be a finite abelian group. We denote by $Z^3(G,\mathbb C^{\ast})$ the group of normalized $3$-cocycles. For any $\omega\in Z^3(G,\mathbb C^{\ast})$ and $g\in G$ we have the map $$\omega_g (x,y)=\frac{\omega(g,x,y)\omega(x,y,g)}{\omega(x,g,y)}$$ Now let $Z^3(G,\mathbb C^{\ast})_{ab}$ denote the set of all normalized $3$-cocycles $\omega$ for which $\omega_g$ is a $2$-coboundary for all $g\in G$, and let $H^3(G,\mathbb C^{\ast}) _{ab}$ be the corresponding set of cohomology classes. It is not hard to check that $H^3(G,\mathbb C^{\ast})_{ab}$ is a subgroup of $H^3(G,\mathbb C^{\ast})$, and your question is asking for an example when it is a proper subgroup.

An easier way to do this is to look for a different description of $H^3(G,\mathbb C^{\ast})_{ab}$. It is not hard to check that $\omega_{g}(x,y)=\omega_g(y,x)$ for all $(x,y)\in G\times G$ iff $\omega_g$ is a $2$-coboundary. So if you define the map $\psi^{\ast}: H^3(G,\mathbb C^{\ast})\to Hom(\bigwedge^3 G,\mathbb C^{\ast})$ $$\psi^{\ast}([\omega])(x,y,z)=\frac{\omega(x,y,z)\omega(y,z,x)\omega(z,x,y)}{\omega(y,x,z)\omega(z,y,x)\omega(x,z,y)}$$ $\psi^{\ast}([\omega])(x,y,z)=\frac{\omega(x,y,z)\omega(y,z,x)\omega(z,x,y)}{\omega(y,x,z)\omega(z,y,x)\omega(x,z,y)}=\frac{\omega_z(x,y)}{\omega_z(y,x)}$$ then $H^3(G,\mathbb C^{\ast})_{ab}$ is precisely the kernel of $\psi^{\ast}$ (Lemma 7.4 in the paper above). Now $\psi^{\ast}$ is surjective so the question becomes: when is $Hom(\bigwedge^3 G,\mathbb C^{\ast})$ non-trivial? This is the case whenever $G$ is the direct sum of at least $3$ cyclic factors, in particular any $(\mathbb Z/n\mathbb Z)^3$ works.

An explicit example in this case is $\omega$ sending $(x,y,z)$ \omega(x,y,z)=\mu^{x_1y_2z_3}$ where $x=(x_1,x_2,x_3)$ etc. to $\mu^{x_1y_2z_3}$, where and $\mu$ is a primitive $n$th root of unity. Then we have $\psi^{\ast}([\omega])(x,y,z)=\mu^{\det(x,y,z)}$ which is non-trivial. In particular $\omega_x$ is non-trivial in $H^2$.
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Following up on what has already been said, this is an explanation taken from "Group cohomology and gauge equivalence of some twisted quantum doubles", by Geoffrey Mason and Siu-Hung Ng.

Let $G$ be a finite abelian group. We denote by $Z^3(G,\mathbb C^{\ast})$ the group of normalized $3$-cocycles. For any $\omega\in Z^3(G,\mathbb C^{\ast})$ and $g\in G$ we have the map $$\omega_g (x,y)=\frac{\omega(g,x,y)\omega(x,y,g)}{\omega(x,g,y)}$$ Now let $Z^3(G,\mathbb C^{\ast})_{ab}$ denote the set of all normalized $3$-cocycles $\omega$ for which $\omega_g$ is a $2$-coboundary for all $g\in G$, and let $H^3(G,\mathbb C^{\ast}) _{ab}$ be the corresponding set of cohomology classes. It is not hard to check that $H^3(G,\mathbb C^{\ast})_{ab}$ is a subgroup of $H^3(G,\mathbb C^{\ast})$, and your question is asking for an example when it is a proper subgroup.

An easier way to do this is to look for a different description of $H^3(G,\mathbb C^{\ast})_{ab}$. It is not hard to check that $\omega_{g}(x,y)=\omega_g(y,x)$ for all $(x,y)\in G\times G$ iff $\omega_g$ is a $2$-coboundary. So if you define the map $\psi^{\ast}: H^3(G,\mathbb C^{\ast})\to Hom(\bigwedge^3 G,\mathbb C^{\ast})$ $$\psi^{\ast}([\omega])(x,y,z)=\frac{\omega(x,y,z)\omega(y,z,x)\omega(z,x,y)}{\omega(y,x,z)\omega(z,y,x)\omega(x,z,y)}$$ then $H^3(G,\mathbb C^{\ast})_{ab}$ is precisely the kernel of $\psi^{\ast}$ (Lemma 7.4 in the paper above). Now $\psi^{\ast}$ is surjective so the question becomes: when is $Hom(\bigwedge^3 G,\mathbb C^{\ast})$ non-trivial? This is the case whenever $G$ is the direct sum of at least $3$ cyclic factors, in particular any $(\mathbb Z/n\mathbb Z)^3$ works.

An explicit example in this case is $\omega$ sending $(x,y,z)$ where $x=(x_1,x_2,x_3)$ etc. to $\mu^{x_1y_2z_3}$, where $\mu$ is a primitive $n$th root of unity. Then we have $\psi^{\ast}([\omega])(x,y,z)=\mu^{\det(x,y,z)}$ which is non-trivial.